derivative of acos(bx+c+d)

解答

\frac{d}{d x} (a cos (b x + c) + d)

- ab sin (b x + c)

求解步骤

\frac{d}{d x} (a cos (b x + c) + d)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (a cos (b x + c)) + \frac{dd}{d x}

\frac{d}{d x} (a cos (b x + c)) = - ab sin (b x + c)

\frac{dd}{d x} = 0

= - ab sin (b x + c) + 0

化简

= - ab sin (b x + c)

a re a 6 x^{2} - 2 x, x^{3} - 6 x + 3

f (x) = 3 x + 1

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2} - 4 x - 5)

\frac{d}{d x} ((x^{2} + 5)^{5})

\frac{d}{d x} (\frac{x ^{2} + 6}{x})